A balance has unequal arms and pans of unequal weight. It is used to weigh two objects of unequal weight. The first object balances against a weight A, when placed in the left pan and against a weight a, when placed in the right pan. The corresponding weights for the second object are B and b. A third object balances against a weight C, when placed in the left pan. What is its true weight?
Solution
The effect of the unequal arms and pans is that if an object of weight x in the left pan balances an object of weight y in the right pan, then x = hy + k for some constants h and k. Thus if the first object has true weight x, then x = hA + k, a = hx + k. So a = h2A + (h+1)k. Similarly, b = h2B + (h+1)k. Subtracting gives h2 = (a - b)/(A - B). and hence (h+1)k = a - h2A = (bA - aB)/(A - B).
The true weight of the third object is thus hC + k = √( (a-b)/(A-B) ) C + (bA - aB)/(A - B) 1/(√( (a-b)/(A-B) ) + 1).
© John Scholes
jscholes@kalva.demon.co.uk
11 May 2002